\[\frac{{\mathrm{d}}^{2}w}{{\mathrm{d}z}^{2}}+\frac{1}{z}\frac{\mathrm{d}w}{%
\mathrm{d}z}+\left(1-\frac{\nu^{2}}{z^{2}}\right)w=\frac{(\tfrac{1}{2}z)^{\nu-%
1}}{\sqrt{\pi}\Gamma\left(\nu+\tfrac{1}{2}\right)}.\]